# Inflection Points

Determine where $f(x)=x^4-8x^3+18x^2$ is concave up and where it is concave down. Find all the inflection points of $f(x) =x^4-8x^3+18x^2$.

A function is concave up where the second derivative is negative and concave down in rest. The inflection points of a function are the points where the second derivative is zero.

$f(x) = x^4-8x^3+18x^2$

$f'(x) = 4x^3-24x^2+36x$

$f”(x) = 12x^2 -48x +36$

$12x^2 -48x +36 = 0$

$x^2 -4x + 3 =0$

$x1 =2+\sqrt{4+3} = 2+\sqrt{7}$

$x2 = 2 -\sqrt{7}$

the second derivative is negative in the interval (x1,x2) and positive in rest. The function is concave up in the interval (x1, x2) and concave down in rest, except the points x1 and x2. The inflection points are x1 and x2.